# Can this situation really be modeled using a Binomial Random Variable? It seems we're sampling without replacement

I've seen the following situation modeled using a Binomial RV:

We know that 20% of the people living in a village are against building an airport while 80% are in favor of it. If we randomly pick a sample of 10 people in the village, what is the probability that 3 of them are against building the airport?

I understand the formula $10 \choose 3$ $(0.2)^3 (0.8)^7$

Here's what I'm confused about: Each trial in a binomial experiment has to have the same probability of success/failure.

Let's say we have our 10 randomly selected people in a room (we don't know yet who they are); we ask one to come out first and it turns out it's one of the people against building the airport.

Doesn't that affect the probability that the rest of the randomly selected people in that room are from either group? After all there's one fewer person in the group against to choose from.

Because we're talking about people that aren't being replaced, it seems to me that we're selecting (choosing) people without replacement, so that the probability is never constant from trial to trial.

In that sense, wouldn't it be incorrect to model this situation using a binomial RV?

• Yes, it would be incorrect to use the binomial RV, but the difference between the exact formula and the approximation (with binomial RV) is very small unless the village is very small. Commented Jul 9, 2016 at 20:22
• @DilipSarwate, thank you. Just to make sure, is my reasoning correct about the idea that we're essentially sampling without replacement? Because there were many times when i thought something was obvious when it comes to probability and I turned out to be completely wrong... Commented Jul 9, 2016 at 20:31
• @DilipSarwate You should post that as an answer. Commented Jul 9, 2016 at 20:31